Problem: Assume that $C$ is a positively oriented, simple, closed curve. Let $R$ be the region enclosed by $C$. Use the normal form of Green's theorem to rewrite $ \iint_R \sin(x + y) - \cos(x - y) \, dA$ as a line integral. Choose 1 answer: Choose 1 answer: (Choice A) A $ \oint_C -8x^3y^2 \, dx + 2\cos(x) \, dy$ (Choice B) B $ \oint_C 2y\sin(x) \, dx - 4x^4y \, dy$ (Choice C) C $ \oint_C -2y\sin(x) \, dx + 4x^4y \, dy$ (Choice D) D $ \oint_C 8x^3y^2 \, dx - 2\cos(x) \, dy$ (Choice E) E Green's theorem is not necessarily applicable.
Assume we have a two-dimensional vector field $F(x, y) = P(x, y) \hat{\imath} + Q(x, y) \hat{\jmath}$ and a positively oriented, piecewise smooth, simple, closed curve $C$. If $R$ is the region enclosed by $C$, then the normal form of Green's theorem states: $ \oint_C -Q \, dx + P \, dy = \iint_R \left( \dfrac{\partial P}{\partial x} + \dfrac{\partial Q}{\partial y} \right) dA$ If $C$ is negatively oriented, then the line integral is equal to the negative of the double integral. [How is this equivalent to the circulation form of Green's thoeorem?] Our first step should be to confirm that the given curve is compatible with using Green's theorem. Looking closely, the curve $C$ does not satisfy all the properties required for Green's theorem: the problem never specifies that $C$ is piecewise-smooth! Because we can't be sure whether or not the curve is piecewise smooth, Green's theorem is not necessarily applicable.